0:13

So now we already have finished our discussion of

Â the black hole type solutions in General Cereal for.

Â In the remaining part of this lecture we want to discuss two complicated phenomena.

Â The first one is, we want to discuss behavior

Â of the electromagnetic waves in the vicinity of the black hole.

Â So in particular, we want to consider the following situation

Â that there is a black hole already created.

Â So this is r=rg.

Â In the very vicinity of this black hole

Â the radius r0 which is rg+epsilon.

Â There is a creation of a wave packet.

Â There is a wave packet created at this position.

Â And climbs out from the gravitational potential and goes to some radius r,

Â which is much greater than r0, which is approximately rg.

Â 1:37

the Full Length Law, because it goes along the radius.

Â So, its equation of motion is given by this relation.

Â O=dS squared=(1-rg/r) dt

Â squared- dr squared /(1- rg/r).

Â From here one can find that time, which takes for

Â this wave packet to climb from this from this position to this position,

Â is equal to the integral from

Â r0 to r of dr/1-(rg/r)).

Â Well that follows from here.

Â This is just equivalent to r-r0+rg.log (r-rg/r0-rg).

Â And now, we use the fact

Â that r is much

Â greater than r0,

Â hence much greater than rg.

Â This is approximately equal to r+rg*log

Â (r/epsilon).

Â So this establishes a relation between Position radius

Â at which, how long does it take?

Â How much time does it take for

Â the wave packet to climb from radius r0 to a fixed radius r?

Â As this guy, this wave packet climbs up, the gravitational potential.

Â It performs a work against

Â gravitational force, hence its energy is decreasing.

Â Because its energy is decreasing,

Â it is natural to expect that its frequency also is decreasing.

Â How its increasing can be seen from the fact that at radius,r,

Â the relation between proper time

Â 4:01

But this time defines for us the frequency, because,

Â omega is proportional to 1/delta t.

Â So the frequency is just, clicking rate.

Â So if this is shorter this is bigger, frequency is bigger.

Â As a result we have the following relation that this guy,

Â omega times

Â g00(r) omega 1 at

Â g00(r)=l to omega 2 at g00(r2).

Â 5:52

As a result, as wave packet climbs up and

Â time goes by, its frequency behaves according to this law,

Â that omega is approximately

Â equal to omega 0*square root of (r/rg)*

Â the exponent of -t-r/(2rg).

Â And as a result the phase of the electric magnetic wave,

Â 6:28

which is equal to the integral from 0 to t dt'

Â omega(t') behaving like this.

Â It is approximately equal to -t2 omega 0 *square

Â root of (r*rg) * exponent

Â of -t -r/(2rg+ con).

Â Irrelevant for further discussion.

Â So this is a behavior of the face, what did we obtain?

Â Let me clarify, what did we obtain.

Â We basically obtained the following fact.

Â 7:25

Which letter should I use?

Â I probably should use k*k squared.

Â So, this is a solution of this equation.

Â As k goes to 0, k goes to 0, this

Â 7:41

wave equation following for Maxwell equation describing some component,

Â either of electromagnetic vector potential or

Â component of electric field or magnetic field.

Â So in this limit as K goes to zero, we obtain so-called Eikonall equation.

Â 8:05

So basically, what we did, we found the solution of

Â this equation in the conditions that we have established,

Â that it started from here and climbed up.

Â So we have obtained this guy, this is what the meaning of all these manipulations.

Â So now one can see what one obtains, where we are?

Â We have obtained the following situation, that

Â basically we implicitly have found the solution

Â of this wave equation in the eikonal approximation.

Â Eikonal, or quasiclassical, or geometric optic approximation.

Â So this is the equation appearing in the geometric optic, where instead of

Â waves like solution of this equation, we obtain straight line.

Â How you say it?

Â 9:33

So, to this wave packet that we are discussing.

Â We want to make it Fourier expansion,

Â which is the integral from 0 to infinity over dt times

Â exponent over here* the exponent

Â of this guy -2i omega

Â 10:24

square root of

Â (r*rg) to the power of (2i

Â omega rg)* exponent of (-pi omega rg)*,

Â eta gamma function (-2i omega g).

Â And in this equation we dropped off all the factors that do not depend on omega,

Â we kept only the factors depending on omega.

Â This is the spectrum that we obtained for

Â the wave packet climbing up

Â from the potential, the geometry.

Â And the spectrum then is just square root of the modulus of this function.

Â It is proportional to the -2 pi

Â omega rg*the gamma function of

Â (-2i omega rg) squared.

Â And this is approximately, well,

Â this is actually equal to pi/omega

Â g (1/ minus exponent

Â of( 4pi omega rg -1).

Â But geometric optic approximation is valid in the limit when omega,

Â when the wavelengths of the packet is much less than rg.

Â Which means that this is true.

Â So only this approximation or this manipulation is correct.

Â This is approximately equal to (pi*omega

Â rg) *exponent (-4pi omega rg).

Â So this all have been obtained in the dramatic optic approximation.

Â We in no way used quantum mechanics anything,

Â so h bar was not present anywhere.

Â But if we restore it there and define that energy as just h bar times omega.

Â Then this guy is proportional

Â to the Boltzmann distribution,

Â where T=h bar/4pi rg.

Â 13:08

So, this is so-called Hawking temperature.

Â Hawking temperature describing Hawking radiation.

Â Of course, Hawking radiation is much more complicated process,

Â it's quantum mechanical.

Â Demands a much deeper study than what we have done.

Â We just observe that if there is a source of radiation in the vicinity

Â of the horizon, due to the geometry, due to the specific Schwarzchild geometry.

Â This actually exponential factor follows from the specific

Â of the Schwarzchild geometry.

Â And due to the specifics of the Schwarzchild geometry,

Â this wave packet as it climbs up from the black hole to infinity, thermalizes.

Â So it behaves as if it is described by the thermal spectrum.

Â Hawking radiation, of course, demands a different derivation.

Â It is due to the change of the ground state of the quantum field theory.

Â So, of course, this is not a proof of the Hawking radiation, but

Â this is a hint why the Hawking radiation should be thermal.

Â If there is something created in the vicinity of the horizon,

Â it will become thermal in the infinity.

Â